$z=-5+3i$ Find the angle $\theta$ (in degrees ) that $z$ makes in the complex plane. Round your answer, if necessary, to the nearest tenth. Express $\theta$ between $\textit{-180}^{\,\circ}$ and $\textit{180}^{\,\circ}$. $\theta=$
Explanation: The strategy We can find the angle $\theta$ of any complex number $z$ by solving the following equation. $\tan\theta=\dfrac{\text{Im}(z)}{\text{Re}(z)}$ This equation usually has two solutions in the interval $[-180,180]$. We can find the appropriate solution by reasoning about the quadrant in which $z$ lies. Solving for $\theta$ $\begin{aligned}\tan\theta &= \dfrac{\text{Im}(z)}{\text{Re}(z)}\\\\ \tan\theta&=\dfrac{3}{-5}\\\\ \theta&=\arctan\left(-\dfrac{3}{5}\right)&\text{Take the arctangent of both sides}\\\\ \theta&\approx-31.0^{\circ}\end{aligned}$ Using the identity $\tan(180+\theta)=\tan(\theta)$, we know that the following is also a solution of the equation. $180^\circ+(-31.0^\circ)=149.0^\circ$ In order to determine which of these two solutions is the angle of $z$, let's take a look at its graphical representation. ${3}$ ${6}$ ${\llap{-}6}$ ${3}$ ${6}$ ${\llap{-}6}$ $Im$ $Re$ $z$ $\theta$ $Re(z)$ $Im(z)$ Since $z$ lies in Quadrant $\text{II}$, its angle must be in the interval $(90,180)$. Therefore, $\theta=149.0^{\circ}$. Summary $\theta=149.0^{\circ}$